POJ3267 The Cow Lexicon(DP+删词)

The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9041 Accepted: 4293

Description

Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the characters 'a'..'z'. Their cowmunication system, based on mooing, is not very accurate; sometimes they hear words that do not make any sense. For instance, Bessie once received a message that said "browndcodw". As it turns out, the intended message was "browncow" and the two letter "d"s were noise from other parts of the barnyard.

The cows want you to help them decipher a received message (also containing only characters in the range 'a'..'z') of length L (2 ≤ L ≤ 300) characters that is a bit garbled. In particular, they know that the message has some extra letters, and they want you to determine the smallest number of letters that must be removed to make the message a sequence of words from the dictionary.

Input

Line 1: Two space-separated integers, respectively: W and L Line 2: L characters (followed by a newline, of course): the received message Lines 3.. W+2: The cows' dictionary, one word per line

Output

Line 1: a single integer that is the smallest number of characters that need to be removed to make the message a sequence of dictionary words.

Sample Input

6 10 browndcodw cow milk white black brown farmer

Sample Output

2

Source

dp[i]表示message从i位置开始需要删除几个字母

dp[i] = dp[i + 1] + 1 // 表示它在i位置并不匹配，删除i位置的字母，

dp[i] = min(dp[i],dp[i + len + t] + t) //对于长度为len的单词，从i开始需要删除t个字母才能完全匹配，i+len+t表示匹配后的下一个,dp[i+len+t]就是匹配成功下一个位置的需要删除的字母数

#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> using namespace std; char dic[605][50]; char mess[300 + 10]; int w,l,dp[605]; int DP(int x, int len, int y) { int j = 1, tot = 0; while(x <= l) { if(mess[x] == dic[y][j]) j++; else tot++; if(j == len + 1) // j==len的时候还没结束，还要匹配len位置，len是最后一个 return tot; x++; } return -1; } int main() { while(scanf("%d%d", &w, &l) != EOF) { scanf("%s", mess + 1); for(int i = 1; i <= w; i++) { scanf("%s", dic[i] + 1); } memset(dp, 0, sizeof(dp)); for(int i = l; i > 0; i--) { dp[i] = dp[i + 1] + 1; for(int j = 1; j <= w; j++) { int len = strlen(dic[j] + 1); int t = DP(i, len, j); if(t != -1) { dp[i] = min(dp[i], dp[i + len + t] + t); } } } printf("%d\n", dp[1]); } return 0; } View Code

 

 

转载于:https://www.cnblogs.com/zhaopAC/p/5171489.html

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